Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 12B Problems/Problem 19"

(→‎Solution)
(→‎Solution 2)
Line 24: Line 24:
  
 
<math>P = \frac{1}{8} + (\frac{1}{8})P</math>
 
<math>P = \frac{1}{8} + (\frac{1}{8})P</math>
 +
 +
Solving for <math>P</math> gives the same answer as above.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2016|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:00, 8 March 2016

Problem

Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?

$\textbf{(A)}\ \frac{1}{8} \qquad \textbf{(B)}\ \frac{1}{7} \qquad \textbf{(C)}\ \frac{1}{6} \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ \frac{1}{3}$

Solution 1

By: dragonfly

We can solve this problem by listing it as an infinite geometric equation. We get that to have the same amount of tosses, they have a $\frac{1}{8}$ chance of getting all heads. Then the next probability is all of them getting tails and then on the second try, they all get heads. The probability of that happening is $\left(\frac{1}{8}\right)^2$.We then get the geometric equation

$x=\frac{1}{8}+\left(\frac{1}{8}\right)^2+\left(\frac{1}{8}\right)^3...$

And then we find that $x$ equals to $\boxed{\textbf{(B)}\ \frac{1}{7}}$ because of the formula of the sum for an infinite series, $\frac{\frac{1}{8}}{1-\frac{1}{8}} = \frac{1}{8}*\frac{8}{7} = \frac{1}{7}$.


Solution 2

Let them "win" if they all flip their coins the same number of times, and the probability that they win is $P$. The probability that they win on their first flip is $\frac{1}{8}$. If they don't win on their first flip, that means they all flipped tails and that their chances of winning have returned to what they were at the beginning. Then we write:

$P = \frac{1}{8} + (\frac{1}{8})P$

Solving for $P$ gives the same answer as above.

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_19&oldid=77472"